Area of Trapezium | AMC-10A, 2018 | Problem 24

We have to find out the area of BGFD.Given that AG is the angle bisector of \(\angle BAC\) ,\(D\) and \(E\) are the mid points of \(AB\) and \(AC\). so we may say that \(DE ||BC\) by mid point theorm…

So clearly BGFD is a Trapezium.now area of the trapezium=\(\frac{1}{2} (BG+DF) \times height betwween DF and BG\)

can you find out the value of \(BG,DF \) and height between them….?

Can you now finish the problem ……….

Let $BC = a$, $BG = x$, $GC = y$, and the length of the perpendicular to $BC$ through $A$ be $h$.

Therefore area of \(\triangle ABC\)=\(\frac{ah}{2}\)=\(120\)………………..(1)

From the angle bisector theorem, we have that\(\frac{50}{x} = \frac{10}{y}\) i.e \(\frac{x}{y}=5\)

Let \(BC\)=\(a\) then \(BG\)=\(\frac{5a}{6}\) and \(DF\)=\(\frac{1}{2 } \times BG\) i.e \(\frac{5a}{12}\)

now can you find out the area of Trapezium ?

can you finish the problem……..

Therefore area of the Trapezium=\(\frac{1}{2} (BG+DF) \times FG\)=\(\frac{1}{2} (\frac{5a}{6}+\frac{5a}{12}) \times \frac{h}{2}\)=\(\frac{ah}{2} \times \frac{15}{24}\)=\(120 \times \frac{15}{24}\)=\(75\) \((from ……..(1))\)